Problem: Simplify; express your answer in exponential form. Assume $y\neq 0, r\neq 0$. $\dfrac{{(y^{5})^{-5}}}{{(y^{5}r^{2})^{-2}}}$
Solution: To start, try working on the numerator and the denominator independently. In the numerator, we have ${y^{5}}$ to the exponent ${-5}$ . Now ${5 \times -5 = -25}$ , so ${(y^{5})^{-5} = y^{-25}}$ In the denominator, we can use the distributive property of exponents. ${(y^{5}r^{2})^{-2} = (y^{5})^{-2}(r^{2})^{-2}}$ Simplify using the same method from the numerator and put the entire equation together. $\dfrac{{(y^{5})^{-5}}}{{(y^{5}r^{2})^{-2}}} = \dfrac{{y^{-25}}}{{y^{-10}r^{-4}}}$ Break up the equation by variable and simplify. $\dfrac{{y^{-25}}}{{y^{-10}r^{-4}}} = \dfrac{{y^{-25}}}{{y^{-10}}} \cdot \dfrac{{1}}{{r^{-4}}} = y^{{-25} - {(-10)}} \cdot r^{- {(-4)}} = y^{-15}r^{4}$.